Mass Flow Rate Uncertainty: A simple problem gone horribly, horribly wrong

Click For Summary
SUMMARY

The discussion centers on calculating the uncertainty in the mass flow rate (Mdot) of a water flow system, determined to be 0.2 kg/s. The user initially calculated the uncertainty as 3.2% but later identified the correct uncertainty as 1.6%. The confusion arose from misinterpreting the uncertainty in mass measurement, which should be ±0.025 kg instead of ±0.05 kg, reflecting the principle that uncertainty is half the smallest increment of the scale. This adjustment significantly impacts the overall uncertainty calculation for the flow rate.

PREREQUISITES
  • Understanding of mass flow rate calculations
  • Familiarity with uncertainty propagation in measurements
  • Knowledge of basic statistics for error analysis
  • Experience with using scales and timing devices in experimental setups
NEXT STEPS
  • Study uncertainty propagation techniques in measurement systems
  • Learn about precision and accuracy in experimental physics
  • Explore the concept of significant figures in data reporting
  • Investigate the use of statistical methods to analyze measurement errors
USEFUL FOR

Students and professionals in engineering, physics, and any field involving experimental measurements and data analysis, particularly those focused on fluid dynamics and measurement uncertainty.

OUmecheng
Messages
18
Reaction score
0
The Mass Flow Rate in a water flow system determined by collecting the discharge over a timed interval is 0.2 kg/s. The scales used can be read to the nearest 0.05kg and stop watch is accurate to 0.2 s. Estimate the precision with which the flow rate can be calculated for time intervals of a.) 10s and b.) 60s

Mdot= mass flow rate

Ok so I found the change in mass by using the flow-rate and given time: m = (Mdot)(change in time) so m = (0.2kg/s)(10s) = 2kg

Then i found the uncertainty in the time and mass:

Ut = 0.2s/10s = 0.02s
Um= 0.05kg/2kg

Then I took uncertainty of the mass flow rate, which came from a bunch of partials like this (d is delta):

UMdot = +/- {[(m/Mdot)(dMdot/delta m)(Um)^2 + (t/Mdot)(dMdot/dt)(Ut)^2]}^(1/2)

Then

UMdot = +/- {[((1)( +/- 0.025)^2 + ((-1)(0.02)^2]}^1/2

=0.032 = 3.2%

BUT the answer is exactly half of that, 1.6%

Where the hell did I go wrong?

I can figure out b.) no problem once I figure out why the initial problem isn't working.
 
Physics news on Phys.org
If the scale can be read to nearest 0.05 kg, what is then the uncertainty in the mass? (hint: it is not ±0.05 kg as that would mean there had to be 0.1 kg between each mark). Same goes for your clock reading.
 
Filip Larsen said:
If the scale can be read to nearest 0.05 kg, what is then the uncertainty in the mass? (hint: it is not ±0.05 kg as that would mean there had to be 0.1 kg between each mark). Same goes for your clock reading.

so half of the smallest increment? Making it ±0.025 kg.
 
OUmecheng said:
so half of the smallest increment? Making it ±0.025 kg.

Indeed. The assumption is that when you read the weight off the scale you select the nearest mass mark and that means you will select a mark no further away from the true reading than half of the distance between the marks. Or, in other words, the true reading will (with some high probability) lie in the range ± 0.025 from the mark.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
36K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 28 ·
Replies
28
Views
8K